Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 1} = \dfrac{4x - 3}{x - 1}$
Explanation: Multiply both sides by $x - 1$ $ \dfrac{x^2}{x - 1} (x - 1) = \dfrac{4x - 3}{x - 1} (x - 1)$ $ x^2 = 4x - 3$ Subtract $4x - 3$ from both sides: $ x^2 - (4x - 3) = 4x - 3 - (4x - 3)$ $ x^2 - 4x + 3 = 0$ Factor the expression: $ (x - 1)(x - 3) = 0$ Therefore $x = 1$ or $x = 3$ At $x = 1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 1$, it is an extraneous solution.